A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car arrives at the position of the stop-light 7.5 s after the light had turned green.
What is the slowest constant speed which the green car can maintain and still catch up to the blue car?
Answer in units of m/s.
To find the slowest constant speed at which the green car can still catch up to the blue car, we need to determine the position of the blue car when the green car arrives at the stop-light.
Let's first find the position of the blue car after 7.5 seconds. We can use the kinematic equation:
d = v_i * t + (1/2) * a * t^2
where:
- d is the final position
- v_i is the initial velocity (which is 0, as the blue car starts from rest)
- a is the acceleration (0.2 m/s^2)
- t is the time (7.5 s)
Substituting the values into the equation, we have:
d = 0 * 7.5 + (1/2) * 0.2 * (7.5)^2
Simplifying, we get:
d = (1/2) * 0.2 * 56.25
d = 5.625 m
So, after 7.5 seconds, the blue car will be 5.625 meters ahead of the stop-light.
Now, let's find the minimum constant speed the green car needs to maintain in order to catch up to the blue car. We can use the equation:
d = v * t
where:
- v is the constant speed of the green car
- d is the distance between the blue car and the stop-light (5.625 m)
- t is the time (7.5 s)
Rearranging the equation to solve for v, we have:
v = d / t
Substituting the known values, we get:
v = 5.625 / 7.5
v ≈ 0.75 m/s
Therefore, the slowest constant speed the green car can maintain and still catch up to the blue car is approximately 0.75 m/s.